@user170039 I suppose one way to think about it is this: if L and R were in fact inverses, we'd have natural equivalences $L\circ R\simeq 1_C$ and $R\circ L\simeq 1_D$. In my experience in category theory, one way to take some kind of structure/data/whatever and weaken it is to replace equivalences with maps just going in one direction or the other (e.g. monoidal to lax monoidal), so maybe this is what people mean by "weak inverse." I don't know.

I'm not sure what the "big picture" behind adjoint functors would be. There are lots of ways, I think, to conceptualize them (e.g. free/forgetful, their relationship to monads, extension/restriction of algebraic structure, "weak mutual inverses" etc.), but I don't have a sense of any of these ways as being some globally "correct" way to think about the structure of adjunctions.

Ultimately, I think, they're useful and that's kind of something that's best understood from a bottom-up, looking at examples point of view.

Is there a closed symmetric monoidal structure on symmetric monoidal $\infty$-categories s.t. the tensor product corepresents bi-symmetric monoidal functors and the enriched hom represents the category of functors with the pointwise symmetric monoidal structure?

@SaalHardali Isn't it just the normal tensor product in the category of commutative algebras? (so it is given at the level of ∞-categories by the cartesian product)

But I don't think that the internal hom (if it exists) has the pointwise symmetric monoidal structure

Oh wait I don't think that either now

Its indepedndent of some things that it should depend on

I agree its the usual tensor product of commutative algebras.

Lets assume we are working in presentable categories for the internal hom question

Well, you need presentably symmetric monoidal categories (i.e. you want the tensor product to commute the colimits in each variable) and I don't think this is true normally for the tensor product of commutative algebras

Yeah i meant $CAlg(Pr^L)$

And yes, in the case of ordinary commutative algebras there's no internal hom

We know it is not true, since it doesn't commute with coproducts

Sorry that was wrong

Aha, that's good to know. Why doesn't it commute with coproducts?

Is there an easy example?

Because the coproduct of commutative algebras is given by the tensor product

I.e. the ⊗ structure on CAlg is the cocartesian one

Aha, i didn'd know this in that generality

It is always true, it's proven in HA in the section where it defines the ⊗-product on CAlg

Cool, thanks!